Equation for plugging in right-invariant vector fields in canonical connection?

Question

Consider a matrix Lie group $G$ with Lie algebra $\frak g$ identified with left-invariant vector fields $\mathcal L(G)$. The $0$-connection is given by: $$ \nabla_{X^l}{Y^l}=\frac{1}{2}[X^l,Y^l]=\frac{1}{2}[X,Y]^l\qquad\forall X^l,Y^l\in\mathcal L(G) $$ where $X=X^l(e),Y=Y^l(e)$ and $e$ is the identity. Note that $[X,Y]$ is simply given by the commutator $XY-YX$.

Now the question is, what seems to be $\nabla_{X^r}{Y^r}$ with $X^r,Y^r$ being right-invariant vector fields? I'm hoping someone could give me a confirmation of my own computation.

1. Since $\nabla_{X^r}Y^r(g_0), g_0\in G$ depends only on the value of $X^r(g_0)$, we can simply treat $X^r(g_0)=Xg_0=g_0(g_0^{-1}Xg_0)$ as the value of $(g_0^{-1}Xg_0)^l\in\mathcal L(G)$ at $g_0$.

2. Next, since $0$-connection implies zero torsion, we have: $$ \nabla_{(g_0^{-1}Xg_0)^l}Y^r-\nabla_{Y^r}(g_0^{-1}Xg_0)^l=[(g_0^{-1}Xg_0)^l,Y^r]=0 $$ $$ \Rightarrow \nabla_{(g_0^{-1}Xg_0)^l}Y^r=\nabla_{Y^r}(g_0^{-1}Xg_0)^l $$ since left and right-variant vector fields commute. Now play the same trick on $Y^r$ and we end up with: $$ \begin{aligned} \nabla_{X^r}Y^r(g_0)&=\nabla_{(g_0^{-1}Yg_0)^l}(g_0^{-1}Xg_0)^l(g_0)\\ &=\frac{1}{2}[(g_0^{-1}Yg_0)^l,(g_0^{-1}Xg_0)^l](g_0)\\ &=\frac{1}{2}[(g_0^{-1}Yg_0),(g_0^{-1}Xg_0)]^l(g_0)\\ &=\frac{1}{2}g_0(g_0^{-1}[Y,X]g_0)\\ &=\frac{1}{2}[Y,X]g_0\\ &=\frac{1}{2}[Y,X]^r(g_0)=\frac{1}{2}[X^r,Y^r](g_0) \end{aligned} $$

Therefore, the conclusion is $\nabla_{X^r}Y^r=\frac{1}{2}[X^r,Y^r]$.